Factoring a third degree polynomial

[Ibraheem]

Homework Statement

Factor out the polynomial and find its solutions x^3-5x^2+7x-12[/B]

Homework Equations

The Attempt at a Solution

I tried to factor it, but I'm stuck in this step x^2(x-5)+7(x-5)+23= 0. I graphed the equation, and I know there is two imaginary solutions and one real positive integer which means there's a linear factor that I can't reach to without graphing the polynomial . So how do I factor the polynomial without referring to the graph? [/B]

 

[ehild]

Try to find the real root among the dividers of the constant term. (what about 4?) It also helps if you find the extrema of the function.

 

[Stephen Tashi]

. So how do I factor the polynomial without referring to the graph?

You could observe the polynomial is negative at x = 0 and (using the way you rewrote it) that it is positive at x = 5. So without graphing it, you can tell the graph crosses the x-axis between x = 0 and x = 5. Look for an integer root that is between 0 and 5 by trial and error.

 

[SteamKing]

Homework Statement

Factor out the polynomial and find its solutions x^3-5x^2+7x-12[/B]

Homework Equations

The Attempt at a Solution

I tried to factor it, but I'm stuck in this step x^2(x-5)+7(x-5)+23= 0. I graphed the equation, and I know there is two imaginary solutions and one real positive integer which means there's a linear factor that I can't reach to without graphing the polynomial . So how do I factor the polynomial without referring to the graph? [/B]

Typically, one uses Descartes rule of signs to determine the number of real roots and the rational root theorem to deduce trial solutions for polynomials.

http://en.wikipedia.org/wiki/Descartes'_rule_of_signs

http://en.wikipedia.org/wiki/Rational_root_theorem

 

[Ray Vickson]

Homework Statement

Factor out the polynomial and find its solutions x^3-5x^2+7x-12[/B]

Homework Equations

The Attempt at a Solution

I tried to factor it, but I'm stuck in this step x^2(x-5)+7(x-5)+23= 0. I graphed the equation, and I know there is two imaginary solutions and one real positive integer which means there's a linear factor that I can't reach to without graphing the polynomial . So how do I factor the polynomial without referring to the graph? [/B]

Use the Rational Root Theorem: see, eg.,
http://en.wikipedia.org/wiki/Rational_root_theorem
or
http://www.purplemath.com/modules/rtnlroot.htm

Note added in edit: I see that the rational root theorem has already been suggested to you.

 

[epenguin]

These problems given with integer coefficients usually have integer solutions (though that's a fact about the problems they give you, not a fact about polynomials with integer coefficients in general). So look at the last term, ±12 or ± one or more of its factors you expect to be a solution. Try. Then there's a bit more to it but which is, or will become, fairly obvious.

 

[haruspex]

Since 12 has so many factors (12, including negatives), trying all of them might be a bit painful. It will certainly help to have an idea of the graph first.
Clearly it is asymptotically -infinity to the left and +infinity to the right.
Differentiating produces a quadratic with obvious factors, so you can find the two local extrema. Necessarily, the one to the left is the local max. But the cubic is still negative there, so you know you are looking for a factor greater than the greater root of the quadratic. There are only four.